# Different solutions under distributive and commutative equivalence

Given 5 numbers: $x_1, x_2, x_3, x_4, x_5 \in \mathbb N$

all the 4 operations: $+ - \times /$

a specific brackets pattern:

$\left(\left(\left(x_1 + x_2 \right) - x_3\right) \times x_4 \right) / x_5 = result$

I can permutate at will the five numbers and the 4 operations, I'm interested in the perumutations that give the same result, so with a equivalence relation such as:

Two permutations that return the same result are said equivalent under the basic mathematic laws: Commutative, Associative and Distributive.

Could be said that two solutions are equivalent if and only if they share the same set of numbers?

For example, the following equations are equivalent:

(((142 + 350) - 372) x 125) / 15 = 1000.0
(((142 + 350) - 372) / 15) x 125 = 1000.0


What I mean is that the brackets are fixed but the numbers and the operator can be moved around at will.

Also, what happens if the condition changes to something like:

$|\left(\left(\left(x_1 + x_2 \right) - x_3\right) \times x_4 \right) / x_5 - result| < \epsilon$

Is the conjecture still Ture/False?

Would this somewhat related to a $\delta$, where $\delta$ might be:

$\delta = min(\delta_{ij}) \quad \forall i,j, \: i \ne j$

with $\delta_{ij} = |x_i - x_j|$

Or nothing could be said?

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I have removed the [calculus-of-variation] and put in [algebra-precalculus] Changes may please be made if irrelevant. – user21436 Feb 12 '12 at 12:02
I removed [algebraic-topology] and added [algorithms]. – Martin Wanvik Feb 12 '12 at 12:09
Not that I know a fast algorithm, but I would suggest to break the problem into two parts: finding all the solutions which result in the desired result, and deciding whether two solutions are equivalent (in your sense) or not. – Tsuyoshi Ito Feb 12 '12 at 19:55
@TsuyoshiIto: The first part is already "solved" (the script I linked does exactly that), the second part is exactly the problem I'm facing: I'm trying to understand if two solutions are equivalent iff they share the same set of numbers, what do you think? – Rik Poggi Feb 12 '12 at 20:01
@RikPoggi: If that is the question, then you should have asked it. It is much simpler than the current question. The answer is no. Consider two formulas (3−2)×6=6 and (6−3)×2=6. If we forget the numbers and just consider them as (x−y)×z and (z−x)×y, they are not equivalent. They happen to be equal because of the particular choice of the numbers. – Tsuyoshi Ito Feb 12 '12 at 20:14